Question

Is this easy derivative question correct?

Answer 1

You don't have to walk me through, I know I need to be doing the work, but I would sure appreciate it if you would tell me if it is wrong so I can re do it.

Answer 2

@welshfella

Answer 3

Anyone have an idea? I can post the formula

Answer 4

Gotta love Patrick JMT on youtube

Answer 5

um um um

Answer 6

lol yes zepdrix? :)

Answer 7

\[\Large\rm f(x)=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So we want to evaluate this at the inverse functions x=2,
which will correspond to the original functions y=2, ya?
\[\Large\rm 2=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So uhhh.. it looks like this is leading to an x value of x=0.

Answer 8

So then uhhhhh.. ya our formula stuff.

Answer 9

Maybe so, looks good?

Answer 10

Ooo ya I made a boo boo somewhere :P
Cause f(0) is = 2.

So we would end up with 1/2.
Woops. Not an option.
Thinkingggg +_+

Answer 11

Its alright I understand this problem is a doozy

Answer 12

Oh oh oh, I'm being silly..
it's not f(0) in the bottom
it's f'(0)

Answer 13

ya ya ya, that should fix things up :)
find your derivative before plugging 0 in

Answer 14

\[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{f^{-1}(2)})}\]
\[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{0})}\]

Answer 15

Eek whats the d dx?

Answer 16

Umm I guess it would be
\[\Large\rm f(x)=0+2x+\sec^2\left(\frac{\pi}{2}x\right)\cdot\frac{\pi}{2}\]I applied chain rule on that last term.

Answer 17

Ok so Im right then?

Answer 18

oh with the blue dot?
ahhh yes, looks correct!! :)

Answer 19

THANK YOU!!!! WHooo hoooo!!!!

Answer 20

yay team \c:/